How do you find modulus and principal value for the argument of #(3 - i)^15#?
2 Answers
Explanation:
First note that:
#abs(3-i) = sqrt(3^2+1^2) = sqrt(10)#
So:
#abs((3-i)^15) = (sqrt(10))^15 = 10^7 sqrt(10)#
I misread the question, so below I have calculated the value of
Note also that we can square a complex number like this:
#(a+bi) = (a^2-b^2) + 2abi#
To calculate the value, I think it may be easiest to square
#(3-i)^15 = (3-i)^16/(3-i)#
#color(white)((3-i)^15) = ((((3-i)^2)^2)^2)^2/(3-i)#
#color(white)((3-i)^15) = (((8-6i)^2)^2)^2/(3-i)#
#color(white)((3-i)^15) = ((28-96i)^2)^2/(3-i)#
#color(white)((3-i)^15) = (-8432-5376i)^2/(3-i)#
#color(white)((3-i)^15) = (42197248+90660864i)/(3-i)#
#color(white)((3-i)^15) = ((42197248+90660864i)(3+i))/((3-i)(3+i))#
#color(white)((3-i)^15) = (35930880 + 314179840i)/10#
#color(white)((3-i)^15) = 3593088 + 31417984i#
Now back to the question as written...
Note that the real and imaginary parts of
Hence the principal value of the argument is simply:
#tan^(-1) (31417984/3593088) = tan^(-1) ((color(red)(cancel(color(black)(128)))*245453)/(color(red)(cancel(color(black)(128)))*28071))= tan^(-1) (245453/28071) ~~ 1.456927#
Explanation:
OK - Let us de Moivre's theorem directly...
#abs(3-i) = sqrt(3^2+1^2) = sqrt(10)#
Hence:
#3-i = sqrt(10)(cos alpha + i sin alpha)#
where
Then:
#(3-i)^15 = (sqrt(10)(cos alpha + i sin alpha))^15#
#color(white)((3-i)^15) = (sqrt(10))^15 (cos 15alpha + i sin 15 alpha)#
#color(white)((3-i)^15) = 10^7sqrt(10) (cos 15alpha + i sin 15 alpha)#
So we want to find the canonical representation of the angle:
#15 tan^(-1)(-1/3) ~~ 15 * (-0.32175) = -4.82625#
Add
#-4.82625+2pi ~~ -4.82625 + 2*3.14159 = 1.45693#